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Positive odd factors

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by kvcpk » Sun Aug 08, 2010 9:23 am
gmatrant wrote:How many positive odd factors does 450 have?
450 = 10 * 45 = 2*5*3^2 * 5 = 2*5^2 * 3^2

Number of factors are (1+1)(2+1)(2+1) = 2*3*3 = 18

Out of these Factors do not contain 2 multiples are odd.

Hence (2+1)(2+1)=9 are odd.

Whats the OA?
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by nicolezl » Sun Aug 08, 2010 9:47 am
prime factorization: 450 = 2 x 3 x 3 x 5 x 5

The odd factors (besides 1) are the odd prime factors and the odd prime factors multiplied by each other (3x3, 5x5, 3x5, 3x3x5, 3x5x5).

positive odd factors of 450: 1, 3, 5, 9, 15, 25, 45, 75

So there are 8 positive odd factors of 450. Is that right?
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by selango » Sun Aug 08, 2010 9:51 am
nicolezl wrote:prime factorization: 450 = 2 x 3 x 3 x 5 x 5

The odd factors (besides 1) are the odd prime factors and the odd prime factors multiplied by each other (3x3, 5x5, 3x5, 3x3x5, 3x5x5).

positive odd factors of 450: 1, 3, 5, 9, 15, 25, 45, 75

So there are 8 positive odd factors of 450. Is that right?
You missed 225[9*25]

9 odd factors.
--Anand--
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by kvcpk » Sun Aug 08, 2010 9:52 am
nicolezl wrote:prime factorization: 450 = 2 x 3 x 3 x 5 x 5

The odd factors (besides 1) are the odd prime factors and the odd prime factors multiplied by each other (3x3, 5x5, 3x5, 3x3x5, 3x5x5).

positive odd factors of 450: 1, 3, 5, 9, 15, 25, 45, 75

So there are 8 positive odd factors of 450. Is that right?
nicolez.. You missed 225. Total turns out to be 9.
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by nicolezl » Sun Aug 08, 2010 10:02 am
kvcpk wrote: nicolez.. You missed 225. Total turns out to be 9.
Ah yes of course! I'm still not sure I understand your reasoning above, even though you did end up with the right answer.
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by kvcpk » Sun Aug 08, 2010 10:08 am
nicolezl wrote:
kvcpk wrote: nicolez.. You missed 225. Total turns out to be 9.
Ah yes of course! I'm still not sure I understand your reasoning above, even though you did end up with the right answer.
Let me explain:

Any number can be written as product of primes.
Example: N = P1^k1 * P2^k2 * P3^k3 *....
Where P1,P2,P3... Are all primes and K1,K2,K3 are their powers.
Then the number of factors of N is given by the formula (k1+1)(k2+1)(k3+1)....

In this example:
450 = 2*5^2 * 3^2
Hence number of factors = (1+1)(2+1)(2+1)= 18
But these incluse both odd and even

Hence to pick up only odd factors, I have ignored powers of 2.
Hence number of odd multiples = (powerof5 +1)(Powerof3 +1)
= (2+1)(2+1)= 9

Hope this helps!!

Let me know in case you have any doubt.
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by nicolezl » Sun Aug 08, 2010 10:13 am
kvcpk wrote:
Let me explain:

Any number can be written as product of primes.
Example: N = P1^k1 * P2^k2 * P3^k3 *....
Where P1,P2,P3... Are all primes and K1,K2,K3 are their powers.
Then the number of factors of N is given by the formula (k1+1)(k2+1)(k3+1)....

In this example:
450 = 2*5^2 * 3^2
Hence number of factors = (1+1)(2+1)(2+1)= 18
But these incluse both odd and even

Hence to pick up only odd factors, I have ignored powers of 2.
Hence number of odd multiples = (powerof5 +1)(Powerof3 +1)
= (2+1)(2+1)= 9

Hope this helps!!

Let me know in case you have any doubt.
Oh ok I get it now, thanks! That's a great rule to know.
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by kvcpk » Sun Aug 08, 2010 10:14 am
nicolezl wrote:
kvcpk wrote:
Let me explain:

Any number can be written as product of primes.
Example: N = P1^k1 * P2^k2 * P3^k3 *....
Where P1,P2,P3... Are all primes and K1,K2,K3 are their powers.
Then the number of factors of N is given by the formula (k1+1)(k2+1)(k3+1)....

In this example:
450 = 2*5^2 * 3^2
Hence number of factors = (1+1)(2+1)(2+1)= 18
But these incluse both odd and even

Hence to pick up only odd factors, I have ignored powers of 2.
Hence number of odd multiples = (powerof5 +1)(Powerof3 +1)
= (2+1)(2+1)= 9

Hope this helps!!

Let me know in case you have any doubt.
Oh ok I get it now, thanks! That's a great rule to know.
You are welcome :)
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